3.3.92 \(\int \frac {a+b x^2+c x^4}{x^2 (d+e x^2)^3} \, dx\) [292]

Optimal. Leaf size=127 \[ -\frac {a}{d^3 x}-\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^2 e \left (d+e x^2\right )^2}+\frac {\left (c d^2+e (3 b d-7 a e)\right ) x}{8 d^3 e \left (d+e x^2\right )}+\frac {\left (c d^2+3 e (b d-5 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} e^{3/2}} \]

[Out]

-a/d^3/x-1/4*(a*e^2-b*d*e+c*d^2)*x/d^2/e/(e*x^2+d)^2+1/8*(c*d^2+e*(-7*a*e+3*b*d))*x/d^3/e/(e*x^2+d)+1/8*(c*d^2
+3*e*(-5*a*e+b*d))*arctan(x*e^(1/2)/d^(1/2))/d^(7/2)/e^(3/2)

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Rubi [A]
time = 0.13, antiderivative size = 124, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1273, 467, 464, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 e (b d-5 a e)+c d^2\right )}{8 d^{7/2} e^{3/2}}-\frac {x \left (\frac {c}{e}-\frac {b d-a e}{d^2}\right )}{4 \left (d+e x^2\right )^2}+\frac {x \left (e (3 b d-7 a e)+c d^2\right )}{8 d^3 e \left (d+e x^2\right )}-\frac {a}{d^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(x^2*(d + e*x^2)^3),x]

[Out]

-(a/(d^3*x)) - ((c/e - (b*d - a*e)/d^2)*x)/(4*(d + e*x^2)^2) + ((c*d^2 + e*(3*b*d - 7*a*e))*x)/(8*d^3*e*(d + e
*x^2)) + ((c*d^2 + 3*e*(b*d - 5*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(7/2)*e^(3/2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 467

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1273

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(m
/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*
p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x],
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rubi steps

\begin {align*} \int \frac {a+b x^2+c x^4}{x^2 \left (d+e x^2\right )^3} \, dx &=-\frac {\left (\frac {c}{e}-\frac {b d-a e}{d^2}\right ) x}{4 \left (d+e x^2\right )^2}-\frac {\int \frac {-4 a d e^2-e \left (c d^2+3 e (b d-a e)\right ) x^2}{x^2 \left (d+e x^2\right )^2} \, dx}{4 d^2 e^2}\\ &=-\frac {\left (\frac {c}{e}-\frac {b d-a e}{d^2}\right ) x}{4 \left (d+e x^2\right )^2}+\frac {\left (c d^2+e (3 b d-7 a e)\right ) x}{8 d^3 e \left (d+e x^2\right )}+\frac {\int \frac {8 a e^2+e \left (c d+e \left (3 b-\frac {7 a e}{d}\right )\right ) x^2}{x^2 \left (d+e x^2\right )} \, dx}{8 d^2 e^2}\\ &=-\frac {a}{d^3 x}-\frac {\left (\frac {c}{e}-\frac {b d-a e}{d^2}\right ) x}{4 \left (d+e x^2\right )^2}+\frac {\left (c d^2+e (3 b d-7 a e)\right ) x}{8 d^3 e \left (d+e x^2\right )}+\frac {\left (c d^2+3 e (b d-5 a e)\right ) \int \frac {1}{d+e x^2} \, dx}{8 d^3 e}\\ &=-\frac {a}{d^3 x}-\frac {\left (\frac {c}{e}-\frac {b d-a e}{d^2}\right ) x}{4 \left (d+e x^2\right )^2}+\frac {\left (c d^2+e (3 b d-7 a e)\right ) x}{8 d^3 e \left (d+e x^2\right )}+\frac {\left (c d^2+3 e (b d-5 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} e^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 124, normalized size = 0.98 \begin {gather*} \frac {\frac {\sqrt {d} \left (-a e \left (8 d^2+25 d e x^2+15 e^2 x^4\right )+d x^2 \left (c d \left (-d+e x^2\right )+b e \left (5 d+3 e x^2\right )\right )\right )}{e x \left (d+e x^2\right )^2}+\frac {\left (c d^2+3 e (b d-5 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}}{8 d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)/(x^2*(d + e*x^2)^3),x]

[Out]

((Sqrt[d]*(-(a*e*(8*d^2 + 25*d*e*x^2 + 15*e^2*x^4)) + d*x^2*(c*d*(-d + e*x^2) + b*e*(5*d + 3*e*x^2))))/(e*x*(d
+ e*x^2)^2) + ((c*d^2 + 3*e*(b*d - 5*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/e^(3/2))/(8*d^(7/2))

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Maple [A]
time = 0.16, size = 111, normalized size = 0.87

method	result	size
default	\(-\frac {\frac {\left (\frac {7}{8} a \,e^{2}-\frac {3}{8} d e b -\frac {1}{8} c \,d^{2}\right ) x^{3}+\frac {d \left (9 a \,e^{2}-5 d e b +c \,d^{2}\right ) x}{8 e}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (15 a \,e^{2}-3 d e b -c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 e \sqrt {d e}}}{d^{3}}-\frac {a}{d^{3} x}\)	\(111\)
risch	\(\frac {-\frac {\left (15 a \,e^{2}-3 d e b -c \,d^{2}\right ) x^{4}}{8 d^{3}}-\frac {\left (25 a \,e^{2}-5 d e b +c \,d^{2}\right ) x^{2}}{8 d^{2} e}-\frac {a}{d}}{x \left (e \,x^{2}+d \right )^{2}}-\frac {15 e \ln \left (-\sqrt {-d e}\, x -d \right ) a}{16 \sqrt {-d e}\, d^{3}}+\frac {3 \ln \left (-\sqrt {-d e}\, x -d \right ) b}{16 \sqrt {-d e}\, d^{2}}+\frac {\ln \left (-\sqrt {-d e}\, x -d \right ) c}{16 \sqrt {-d e}\, e d}+\frac {15 e \ln \left (-\sqrt {-d e}\, x +d \right ) a}{16 \sqrt {-d e}\, d^{3}}-\frac {3 \ln \left (-\sqrt {-d e}\, x +d \right ) b}{16 \sqrt {-d e}\, d^{2}}-\frac {\ln \left (-\sqrt {-d e}\, x +d \right ) c}{16 \sqrt {-d e}\, e d}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)/x^2/(e*x^2+d)^3,x,method=_RETURNVERBOSE)

[Out]

-1/d^3*(((7/8*a*e^2-3/8*d*e*b-1/8*c*d^2)*x^3+1/8*d*(9*a*e^2-5*b*d*e+c*d^2)/e*x)/(e*x^2+d)^2+1/8*(15*a*e^2-3*b*
d*e-c*d^2)/e/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2)))-a/d^3/x

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Maxima [A]
time = 0.52, size = 121, normalized size = 0.95 \begin {gather*} \frac {{\left (c d^{2} e + 3 \, b d e^{2} - 15 \, a e^{3}\right )} x^{4} - 8 \, a d^{2} e - {\left (c d^{3} - 5 \, b d^{2} e + 25 \, a d e^{2}\right )} x^{2}}{8 \, {\left (d^{3} x^{5} e^{3} + 2 \, d^{4} x^{3} e^{2} + d^{5} x e\right )}} + \frac {{\left (c d^{2} + 3 \, b d e - 15 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {3}{2}\right )}}{8 \, d^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^2/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

1/8*((c*d^2*e + 3*b*d*e^2 - 15*a*e^3)*x^4 - 8*a*d^2*e - (c*d^3 - 5*b*d^2*e + 25*a*d*e^2)*x^2)/(d^3*x^5*e^3 + 2
*d^4*x^3*e^2 + d^5*x*e) + 1/8*(c*d^2 + 3*b*d*e - 15*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-3/2)/d^(7/2)

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Fricas [A]
time = 0.38, size = 428, normalized size = 3.37 \begin {gather*} \left [-\frac {2 \, c d^{4} x^{2} e + 30 \, a d x^{4} e^{4} + {\left (15 \, a x^{5} e^{4} - c d^{4} x - 3 \, {\left (b d x^{5} - 10 \, a d x^{3}\right )} e^{3} - {\left (c d^{2} x^{5} + 6 \, b d^{2} x^{3} - 15 \, a d^{2} x\right )} e^{2} - {\left (2 \, c d^{3} x^{3} + 3 \, b d^{3} x\right )} e\right )} \sqrt {-d e} \log \left (\frac {x^{2} e + 2 \, \sqrt {-d e} x - d}{x^{2} e + d}\right ) - 2 \, {\left (3 \, b d^{2} x^{4} - 25 \, a d^{2} x^{2}\right )} e^{3} - 2 \, {\left (c d^{3} x^{4} + 5 \, b d^{3} x^{2} - 8 \, a d^{3}\right )} e^{2}}{16 \, {\left (d^{4} x^{5} e^{4} + 2 \, d^{5} x^{3} e^{3} + d^{6} x e^{2}\right )}}, -\frac {c d^{4} x^{2} e + 15 \, a d x^{4} e^{4} + {\left (15 \, a x^{5} e^{4} - c d^{4} x - 3 \, {\left (b d x^{5} - 10 \, a d x^{3}\right )} e^{3} - {\left (c d^{2} x^{5} + 6 \, b d^{2} x^{3} - 15 \, a d^{2} x\right )} e^{2} - {\left (2 \, c d^{3} x^{3} + 3 \, b d^{3} x\right )} e\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}} - {\left (3 \, b d^{2} x^{4} - 25 \, a d^{2} x^{2}\right )} e^{3} - {\left (c d^{3} x^{4} + 5 \, b d^{3} x^{2} - 8 \, a d^{3}\right )} e^{2}}{8 \, {\left (d^{4} x^{5} e^{4} + 2 \, d^{5} x^{3} e^{3} + d^{6} x e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^2/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

[-1/16*(2*c*d^4*x^2*e + 30*a*d*x^4*e^4 + (15*a*x^5*e^4 - c*d^4*x - 3*(b*d*x^5 - 10*a*d*x^3)*e^3 - (c*d^2*x^5 +
6*b*d^2*x^3 - 15*a*d^2*x)*e^2 - (2*c*d^3*x^3 + 3*b*d^3*x)*e)*sqrt(-d*e)*log((x^2*e + 2*sqrt(-d*e)*x - d)/(x^2
*e + d)) - 2*(3*b*d^2*x^4 - 25*a*d^2*x^2)*e^3 - 2*(c*d^3*x^4 + 5*b*d^3*x^2 - 8*a*d^3)*e^2)/(d^4*x^5*e^4 + 2*d^
5*x^3*e^3 + d^6*x*e^2), -1/8*(c*d^4*x^2*e + 15*a*d*x^4*e^4 + (15*a*x^5*e^4 - c*d^4*x - 3*(b*d*x^5 - 10*a*d*x^3
)*e^3 - (c*d^2*x^5 + 6*b*d^2*x^3 - 15*a*d^2*x)*e^2 - (2*c*d^3*x^3 + 3*b*d^3*x)*e)*sqrt(d)*arctan(x*e^(1/2)/sqr
t(d))*e^(1/2) - (3*b*d^2*x^4 - 25*a*d^2*x^2)*e^3 - (c*d^3*x^4 + 5*b*d^3*x^2 - 8*a*d^3)*e^2)/(d^4*x^5*e^4 + 2*d
^5*x^3*e^3 + d^6*x*e^2)]

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Sympy [A]
time = 1.10, size = 202, normalized size = 1.59 \begin {gather*} \frac {\sqrt {- \frac {1}{d^{7} e^{3}}} \cdot \left (15 a e^{2} - 3 b d e - c d^{2}\right ) \log {\left (- d^{4} e \sqrt {- \frac {1}{d^{7} e^{3}}} + x \right )}}{16} - \frac {\sqrt {- \frac {1}{d^{7} e^{3}}} \cdot \left (15 a e^{2} - 3 b d e - c d^{2}\right ) \log {\left (d^{4} e \sqrt {- \frac {1}{d^{7} e^{3}}} + x \right )}}{16} + \frac {- 8 a d^{2} e + x^{4} \left (- 15 a e^{3} + 3 b d e^{2} + c d^{2} e\right ) + x^{2} \left (- 25 a d e^{2} + 5 b d^{2} e - c d^{3}\right )}{8 d^{5} e x + 16 d^{4} e^{2} x^{3} + 8 d^{3} e^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)/x**2/(e*x**2+d)**3,x)

[Out]

sqrt(-1/(d**7*e**3))*(15*a*e**2 - 3*b*d*e - c*d**2)*log(-d**4*e*sqrt(-1/(d**7*e**3)) + x)/16 - sqrt(-1/(d**7*e
**3))*(15*a*e**2 - 3*b*d*e - c*d**2)*log(d**4*e*sqrt(-1/(d**7*e**3)) + x)/16 + (-8*a*d**2*e + x**4*(-15*a*e**3
+ 3*b*d*e**2 + c*d**2*e) + x**2*(-25*a*d*e**2 + 5*b*d**2*e - c*d**3))/(8*d**5*e*x + 16*d**4*e**2*x**3 + 8*d**
3*e**3*x**5)

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Giac [A]
time = 3.30, size = 110, normalized size = 0.87 \begin {gather*} \frac {{\left (c d^{2} + 3 \, b d e - 15 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {3}{2}\right )}}{8 \, d^{\frac {7}{2}}} - \frac {a}{d^{3} x} + \frac {{\left (c d^{2} x^{3} e + 3 \, b d x^{3} e^{2} - c d^{3} x - 7 \, a x^{3} e^{3} + 5 \, b d^{2} x e - 9 \, a d x e^{2}\right )} e^{\left (-1\right )}}{8 \, {\left (x^{2} e + d\right )}^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^2/(e*x^2+d)^3,x, algorithm="giac")

[Out]

1/8*(c*d^2 + 3*b*d*e - 15*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-3/2)/d^(7/2) - a/(d^3*x) + 1/8*(c*d^2*x^3*e + 3
*b*d*x^3*e^2 - c*d^3*x - 7*a*x^3*e^3 + 5*b*d^2*x*e - 9*a*d*x*e^2)*e^(-1)/((x^2*e + d)^2*d^3)

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Mupad [B]
time = 0.39, size = 118, normalized size = 0.93 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2+3\,b\,d\,e-15\,a\,e^2\right )}{8\,d^{7/2}\,e^{3/2}}-\frac {\frac {a}{d}-\frac {x^4\,\left (c\,d^2+3\,b\,d\,e-15\,a\,e^2\right )}{8\,d^3}+\frac {x^2\,\left (c\,d^2-5\,b\,d\,e+25\,a\,e^2\right )}{8\,d^2\,e}}{d^2\,x+2\,d\,e\,x^3+e^2\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)/(x^2*(d + e*x^2)^3),x)

[Out]

(atan((e^(1/2)*x)/d^(1/2))*(c*d^2 - 15*a*e^2 + 3*b*d*e))/(8*d^(7/2)*e^(3/2)) - (a/d - (x^4*(c*d^2 - 15*a*e^2 +
3*b*d*e))/(8*d^3) + (x^2*(25*a*e^2 + c*d^2 - 5*b*d*e))/(8*d^2*e))/(d^2*x + e^2*x^5 + 2*d*e*x^3)

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